Let $d:\mathbb{N}\rightarrow\mathbb{N}$ where$d(n)$ is the number of natural divisors of $n$. Prove that for all $n$, $d(2^{n-1})=n$. (I formulated this conjecture myself. So, if it's nonsensical let me know.)

Induction:

Base: let $n=1$. Then $d(2^{1-1})=1$.

Now assume for some $k\in\mathbb{N}$, $d(2^{k-1})=k$. Then the number of natural divisors of $2^{k-1}$ can be listed as follows:

$2^0,2^1,2^2,...,2^{k-1}$. Multiplying each divisor by 2 gives the list $2^1,2^2,..,2^{k-1},2^{(k+1)-1}$. But these are the divisors of $2^{(k+1)-1}$ without $1$. Thus $d(2^{(k+1)-1})=d(2^{k-1})+1=k+1$ as desired.

Therefore by the principle of induction, $d(2^{k-1})=k$.